Using the formula \( v = \sqrt{\dfrac{\gamma P}{\rho }}ργP\)​​  explain why the speed of sound in air

(a) does not depend upon pressure.

(b) increases with temperature and humidity.

(c) increases with humidity.

Asked by Pragya Singh | 1 year ago |  120

1 Answer

Solution :-


v = \( \sqrt{\dfrac{\gamma P}{\rho }}ργP​​\)

We know,

PV = nRT ( for n moles of ideal gas )

= PV = RT (\( \dfrac{m}{M}\))

Where, m is the total mass and M is molecular mass of the gas.

therefore, P 

 = m (\( \dfrac{RT}{M}\))

= P = \( \dfrac{\rho RT}{M}MρRT​\)

\( \dfrac{P}{\rho} = \dfrac{ RT}{M}​=MRT​\)

(a) For a gas at a constant temperature,

\( \dfrac{P}{\rho}\) = constant

Thus, as P increases ρ and vice versa. 

This means that \( \dfrac{P}{\rho}\) ratio always remains constant meaning

v =\( \sqrt{\dfrac{\gamma P}{\rho }}ργP​​\)
constant i.e., velocity of sound does not depend upon the pressure of the gas.


( b ) Since, \( \dfrac{P}{\rho}\)

\( = \dfrac{ RT}{M}ρP​=MRT​\)

v =  \( \sqrt{\dfrac{\gamma P}{\rho }}ργP​​\)

 v =\( \sqrt{\dfrac{\gamma RT}{M }}MγRT​​\)

We can see that v \( \propto \sqrt{T}∝T\)​ i.e., speed of sound increases with temperature.


(c) When humidity increases, effective density of the air decrease. 

This means v \( \propto \dfrac{1}{\sqrt{\rho }}v∝ρ​1\)​, thus velocity increases.

Answered by Abhisek | 1 year ago

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